What substances are acidic

2 answers:

7 0

A substance with LOW pH. The lower the pH, the higher the acidity level is. pH has a direct effect on how acidic a substance is. For an example, battery acid. Battery acid has a pH level of about 1. This means that battery acid is very acidic. A substance is concidered acidic when it dips below the neutral pH level of 7.

Alex_Xolod [135]3 years ago

4 0

Answer:

PH7 is the normal level of about tap water/normal water.If anything is below PH7 then it is acidic,You would then call it LOW PH level.

Explanation:

You might be interested in

Why do waves with high frequencies have short wavelengths

dalvyx [7]

Answer: because ν = velocity/λ where ν and λ are the frequency and wavelegth of the wave.

Explanation: In order to explain this problem we have to consider the relationship between frequency and wavelengths which are related by the velocity of the wave as follows ν*λ=v where ν and λ are the frequency and wavelegth of the wave. These parameters have an inverse proportionality.

Then, ν = velocity/λ

7 0

3 years ago

Read 2 more answers

A 6000 kg roller coaster goes around a loop of radius 30 m at 6 m/s. What is the centripetal acceleration?

Margarita [4]

Answer:

The answer to the question is 7200

7 0

3 years ago

A 70.0-kg person throws a 0.0480-kg snowball forward with a ground speed of 33.5 m/s. A second person, with a mass of 55.0 kg, c

saw5 [17]

Answer:

The final velocity of the thrower is $\bf{3.88~m/s}$ and the final velocity of the catcher is $\bf{0.029~m/s}$ .

Explanation:

Given:

The mass of the thrower, $m_{t} = 70~Kg$ .

The mass of the catcher, $m_{c} = 55~Kg$ .

The mass of the ball, $m_{b} = 0.0480~Kg$ .

Initial velocity of the thrower, $v_{it} = 3.90~m/s$

Final velocity of the ball, $v_{fb} = 33.5~m/s$

Initial velocity of the catcher, $v_{ic} = 0~m/s$

Consider that the final velocity of the thrower is $v_{ft}$ . From the conservation of momentum,

$&& m_{t}v_{ft} + m_{b}v_{fb} = (m_{t} + m_{b})v_{it}\\&or,& v_{ft} = \dfrac{(m_{t} + m_{b})v_{it} - m_{b}v_{fb}}{m_{t}}\\&or,& v_{ft} = \dfrac{(70 + 0.0480)(3.90) - (0.0480)(33.5)}{70}\\&or,& v_{ft} = 3.88~m/s$

Consider that the final velocity of the catcher is $v_{fc}$ . From the conservation of momentum,

$&& (m_{c} + m_{b})v_{fc} = m_{b}v_{it}\\&or,& v_{fc} = \dfrac{m_{b}v_{it}}{(m_{c} + m_{b})}\\&or,& v_{fc} = \dfrac{(0.048)(33.5)}{(55.0 + 0.0480)}\\&or,& v_{fc} = 0.029~m/s$